Mumbai: Indian cricket team coach Ravi Shastri said that Washington Sundar’s performance for Royal Challengers Bangalore (RCB) against Mumbai Indians is the best in the current Indian Premier League (IPL) 2020.
His remark came as RCB defeated Mumbai Indians in the Super Over on Monday at the Dubai International Stadium.
For RCB, Sundar conceded 12 runs in his quota of just four overs and also took the crucial wicket of Rohit Sharma.
Taking note of Sundar’s performance, Shastri tweeted: “In a batsman’s world – from Chennai to Washington. Best IPL performance so far in 2020. Special.”
— Ravi Shastri (@RaviShastriOfc) September 28, 2020
For Royal Challengers Bangalore opener Devdutt Padikkal played another brilliant knock of 53 runs of just 37 balls whereas Ab De Villiers had played an unbeaten knock of 55 runs off just 24 balls with the help of 4 fours and 4 sixes to take RCB’s score to 201/3 in the allotted twenty overs.
For Mumbai Indians, it was not a good start as they lost their top three batsmen at an early stage of the innings. Ishan Kishan played a knock of 99, studded with two fours and nine sixes.
On the other hand, Pollard registered an unbeaten inning of 60 off just 24 balls with the help of three fours and five sixes. With this 119 runs to take the side closer to the target.
In the Super Over, Mumbai Indians could manage only seven runs as Saini bowled a brilliant over, and was easily chased by Virat Kohli and AB de Villiers. In normal match action, both Mumbai and Bangalore scored 201 runs on the board.
RCB has played three matches in the tournament so far and the side has managed to win two out of these three. The Virat Kohli-led side will next lock horns with Rajasthan Royals on October 3.